what is the grey area ? take strip perpendicular to x-axis what are the limits of integration
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What is the grey area ?
Take strip perpendicular to x-axis
What are the limits of integration.
(1 ) 02
xA dx
2
0(1 )
2
xA dx
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What is the grey area ?
Take strip parallel to x-axis
What are the limits of integration.
(2 2 ) 0A y dy
1
0(2 2 ) 0A y dy
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What is the volumeif the grey area is revolved about the
x-axis?
What are the limits of integration.
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What is the volumeif the grey area is revolved about the
x-axis?
What are the limits of integration.
2 2
0(1 )
2
xV dx
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What is the volumeif the yellow area is revolved about the
y-axis?
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What is the volumeif the yellow area is revolved about the
y-axis?2 2
0
3( )2
yV dy
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When finding the area, use
A. True
B. False
2
0( ) ( )top x bottom x dx
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When finding the area, use
A. True
B. False
2
0( ) ( )top x bottom x dx
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When finding the area, use
A. True
B. False
3
0
2(2 )
3x dx
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When finding the area, use
A. True
B. False
3
0
2(2 )
3x dx
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Revolve about the y-axisV =
A. True
B. False
3
0
22 (2 )
3x x dx
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Revolve about the y-axisV =
A. True
B. False
3
0
22 (2 )
3x x dx
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Revolve about the x-axisV =
A. True
B. False
3 2
0
4(4 )
9x dx
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Revolve about the x-axisV =
A. True
B. False
3 2
0
4(4 )
9x dx
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What is the volumeif the grey area is revolved about the
x-axis?Red strip perpendicular to axis
Solve for y
and square
12
xy
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What is the volumeif the grey area is revolved about the
x-axis?
What are the limits of integration.
2 2
0(1 )
2
xV dx
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What is the volumeif the yellow area is revolved about the
y-axis?
Solve for x, square, integrate, times pi
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What is the volumeif the yellow area is revolved about the
y-axis?
Solve for x, square, integrate, times pi
2 2
0
3( )2
yV dy
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#56 Plumb bob design
Revolve the shown region about the x-axis
23612
xy x
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#56 Plumb bob design
Revolve the shown region about the x-axis
Must weigh in the neighborhood of 190 g.
Specify a brass that weighs 8.5 g/cm3.
23612
xy x
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#56 Plumb bob design
V =
Must weigh in the neighborhood of 190 g.
Specify a brass that weighs 8.5 g/cm3.
23612
xy x
22(36 )
144
xx dx
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#56 Plumb bob design
V =
=22.62
cm3
times 8.5 g/cm3.
=192.27 g.
23612
xy x
26 2
0(36 )
144
xx dx
6 2 4
036
144x x dx
3 5 56 3
0
36 6( ) (12(6 ) )
144 3 5 144 5|x x
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.2 2
0 02 sin cos sin 2V x x dx x dx
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.
3.14159
0.1
2 2
0 02 sin cos sin 2V x x dx x dx
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What is the volumeif the yellow area is rotated about the
y-axis?
Last time we went about the x as shown.
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Dr. Jack Tenzel has a Project-o-Chart in his office
• The light reflects off of a mirror and ends up on a wall in front of the patient.
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Given the center light source, calculate the volume around it
• First write the equation of the surface.
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y = 16 x-4
We will come
back to this later.
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What is the volume of a coke can?
• Just the aluminum
• Top - Bottom
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The volume of a can is 2 r times the height times
the x.
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Start like we did for area. Take a narrow x red strip and then
rotate it about the y-axis.
• This makes a red coke can.• The volume of one can is…• 2x(f(x)-g(x))x so the desired volume is
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Set up n rectangles of width x
Revolve about the y-axis
That produces n cylinders
* *
1
lim 2 ( ( ) ( ) i
n
i in
i
Volume x f x x xg
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Take a narrow x red strip and then rotate it about the y-axis.
• This makes a coke can.
• The volume of one can with radius x is…
• 2x(f(x)-g(x))x so the desired volume is
* *
1
lim 2 ( ( ) ( ) i
n
i in
i
Volume x f x x xg
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By the definition
• Volume = * *
1
lim 2 ( ( ) ( )n
i i in
i
x f x g x x
2 ( ( ) ( ))
b
a
x f x g x dx
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Example 1
• Find the volume when the area under y = x2 and over the x-axis is revolved about the y-axis.
• Between x=0 and x=2• Just add up all of the red coke cans• As they slide from x=0 to x=2• Top function is y= x2
• Bottom function is y = 0
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Example 1• Find the volume when the area under
y=x2
• Between x=0 and x=2• Is revolved about the y-axis
• = 2x4/4
• = 2/4 = 8
22
0
2 ( 0)Volume x x dx 20|
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.
• Back to the problem
• x is the radius times top - bottom
2 ( ( ) ( ))b
a
V x f x g x dx
24
1
2 (16 1)V x x dx
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.2
4
1
2 (16 1)V x x dx 2
3
1
2 (16 )x x dx 2 22 2 2
21 1
| 8 |2 (16 ) 2 ( )
2 2 | 2 |
x x x
x
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A. 2B. 2C. 2
22
21
8 |2 ( )
2 |
x
x
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A. 2B. 2C. 2
22
21
8 |2 ( )
2 |
x
x
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2
A. 2[ - 4 + 8.5]
B. 2[ - 4 - 8.5]
C. 2[ - 4 + 7.5 ]
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2
A. 2[ - 4 + 8.5]
B. 2[ - 4 - 8.5]
C. 2[ - 4 + 7.5 ]
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Example 3
• Revolve the area between x2 and x3 about the y-axis• Find the volume generated.• 0 = x2 ( x – 1 ) • so x2 =0 or x–1=0• Next we add up all of the red cylinders• From 0 to 1• Volume =
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Revolve about the y-axisA. [
B. [
C. [
12 3
0
. 2 [ ]A V x x x dx 2
2 3
0
. 2 [ ]B V x x x dx 1
3 2
0
. 2 [ ]C V x x x dx
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Revolve about the y-axisA. [
B. [
C. [
12 3
0
. 2 [ ]A V x x x dx 2
2 3
0
. 2 [ ]B V x x x dx 1
3 2
0
. 2 [ ]C V x x x dx
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.
A. .
B. .
C. .
12 3 4
0
|. 2 ( [ ])
2 3 4 |
x x xA
14 5
0
|. 2 [ ]
4 5 |
x xB V
12 3
0
|. 2 [3 4 ]
|C V x x
12 3
0
2 [ ]V x x x dx
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.
A. .
B. .
C. .
12 3 4
0
|. 2 ( [ ])
2 3 4 |
x x xA
14 5
0
|. 2 [ ]
4 5 |
x xB V
12 3
0
|. 2 [3 4 ]
|C V x x
12 3
0
2 [ ]V x x x dx
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Volume =
= 2 [(5-4)/20] = 2 /20 = /10
12 3
0
2 ( )V x x x dx
4 510
1 12 [ ] | 2 [ ]
4 5 4 5
x x
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Example 2
• Consider the first region in the first quadrant bounded by y=sin(x2) and y=1/root(2)
• Set the two functions equal• Solve for x2 and then for x• Spin about the y-axis or radius of x• Add the volumes of the n cylinders
* *2
1
22 lim (sin( ) )
2
n
i i ini
V x x x
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Example 2
• sin(x2) = root(2)/2 when x2=/4 or 3/4
* *2
1
22 lim (sin( ) )
2
n
i i ini
V x x x
3
22
2
22 (sin( ) )
2V x x dx
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Example 2
3
22
2
22 (sin( ) )
2V x x dx
3
22
2
(sin( )2 2 )V x x x dx
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Example 22 3
2 2
2
[ cos( ) 2 ] |2
xx
3
22
2
sin( )2 2V x x x dx
3 3 2 2cos cos
4 8 4 8
2 3 2 2 2
2 8 2 8
2
24
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y = x + 3 .…….y = x2 + 1 . …1
• 2 pi radius ( top – bottom) thick.
2 2
12 ( 1)([ 3] [ 1])V x x x dx
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.
A. .
B. .
C. .
2 2
12 ( 1)([ 3] [ 1])V x x x dx
2 3
12 3 2x x dx
2 3
12 3 4x x dx
2 3 2
12 3 4x x dx
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.
A. .
B. .
C. .
2 2
12 ( 1)([ 3] [ 1])V x x x dx
2 3
12 3 2x x dx
2 3
12 3 4x x dx
2 3 2
12 3 4x x dx
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Region of y=x2+1 and y=x+3 is revolved about the line x = -1.
2 2
12 ( 1)([ 3] [ 1])V x x x dx
2 3
12 3 2x x dx
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Region of y=x2+1 and y=x+3 is revolved about the line x = -1.
45.95
0.2
2 2
12 ( 1)([ 3] [ 1])V x x x dx
2 3
12 3 2x x dx
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Region bounded by y=x2+1 and y=x+3 is revolved about the x-axis.
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Region bounded by y=x2+1 and y=x+3 is revolved about the x-axis.
73.51
0.2